Some more electromagnetism questions (ac generators)

[KatieKangaroo]

A few more questions that i am stuck with:
An AC generator has a coil area of 120cm². The coil has 100 turns. The generator produces a peak voltage of 29V when the coil is turning at 6000 revolutions per minute.
Calculate the magnetic flux density (assumed uniform) required.

The rate of rotation is reduced to 3600 revolutions per minute.
Calculate the new peak output voltage and the new output frequency. (I calculated the old frequency before and found that it was 100Hz).

A wire is bent into the shape of a semicircle of radius 0.2m. A uniform magnetic field which has a flux density of 0.8T acts normally to the plane of the semicircular wire. As the wire is rotated at a constant rate it generates an alternating EMF with a peak value between the terminals 0f 0.31V Calculate the angular speed at which the wire is rotated.

Ive been trying to use the formula EMF = ωNABsinθ but without success.
Sorry, I am just having trouble answering these without a teacher to help
Thanks again for any help in advance.

 

[mezarashi]

The formula for EMF you are using seems reasonable. Maybe it's your math, so if you could write out some of your calculations.

 

[quicknote]

Hello, thank you for reaching out with your questions about AC generators and electromagnetism. I am happy to assist you in understanding these concepts.

To calculate the magnetic flux density required for the given AC generator, we can use the equation B = Vpeak / (NACω), where B is the magnetic flux density, Vpeak is the peak voltage, N is the number of turns, A is the coil area, and ω is the angular velocity. Plugging in the given values, we get B = (29V) / (100 * 120cm² * 6000rpm * (2π/60)), which simplifies to B = 0.0000121 Tesla.

When the rate of rotation is reduced to 3600rpm, the new peak output voltage can be calculated using the equation Vpeak = NABω, where Vpeak is the peak voltage, N is the number of turns, A is the coil area, B is the magnetic flux density, and ω is the angular velocity. Plugging in the new values, we get Vpeak = (100 * 120cm² * 0.0000121T * (2π/60) * 3600rpm), which simplifies to Vpeak = 0.87V. The new output frequency can be calculated using the equation f = ω / (2π), where f is the frequency and ω is the angular velocity. Plugging in the new value for ω, we get f = (3600rpm * (2π/60)) / (2π), which simplifies to f = 60Hz.

For the third question, the formula you are using, EMF = ωNABsinθ, is correct. To solve for the angular speed, we can rearrange the equation to ω = EMF / (NABsinθ). Plugging in the given values, we get ω = (0.31V) / (N * 0.2m * 0.8T * sin(180°)), which simplifies to ω = 1.93 rad/s.

I hope this helps you in understanding these concepts better. Keep exploring and asking questions, and remember, there is no shame in seeking help when needed. Good luck with your studies!